A liquid flowing from a vertical pipe has a very definite shape as it flows from the pipe. To get the equation for this shape, assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed v0 and the radius of the stream of liquid is r0. Find an equation for the speed of the liquid as a function of the distance y it has fallen. Express your answer in terms of the given quantities and appropriate constants. Combining this with the equation of continuity, find an expression for the radius of the stream as a function of y. Express your

Question

A liquid flowing from a vertical pipe has a very definite shape as it flows from the pipe. To get the equation for this shape, assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed v0 and the radius of the stream of liquid is r0.       Find an equation for the speed of the liquid as a function of the distance y it has fallen. Express your answer in terms of the given quantities and appropriate constants.     Combining this with the equation of continuity, find an expression for the radius of the stream as a function of y. Express your

anonymous · Answer

Part C question is If water flows out of a vertical pipe at a speed of 1.20 m/s, how far below the outlet will the radius be one-half the original radius of the stream?
Express your answer using three significant figures.

anonymous · Answer

Q is volume rate of flow = (velocity, v) (cross-sectional area, pi r^2) 

Get velocity of falling material from  v^2 = v0^2 + 2 g s,
where v0 is initial velocity, g is gravitational acceleration, s is distance fallen.

With v thus obtained, get r from r^2 = Q / pi v.

anonymous · Answer

When you said "With v thus obtained, get r from r^2 = Q / pi v."   What if we are not supposed to use Q in the answer? but just what we were given